# How to solve $x$ in $(x+1)^4+(x-1)^4=16$?

I'm trying my hand on these types of expressions. How to solve $x$ in $(x+1)^4+(x-1)^4=16$? Please write any idea you have, and try to keep it simple. Thanks.

We have our equation to solve for: $$(x+1)^4+(x-1)^4=16$$ First, expand the terms in the left hand side. $$x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1=16$$ Simplify the left hand side. $$2x^4+12x^2+2=16$$ Factor $2$ out from both sides. $$x^4+6x^2+1=8$$ Move $8$ to the left hand side. $$x^4+6x^2-7=0$$ Let $x^2=a$. Now our equation becomes: $$a^2+6a-7=0$$ Factor it. $$(a-1)(a+7)=0$$ Reverse the substitution. $$(x^2-1)(x^2+7)=0$$ Factor some more (yes even the sum of squares) $$(x+1)(x-1)(x+\sqrt7i)(x-\sqrt7i)=0$$ The four roots are $x=\pm 1$ and $x=\pm \sqrt{7}i$. But if you do not want complex roots, then the two real roots are $x=1$, $x=-1$.

Hope I helped!

• The left hand side should simplify to $2x^4+12x^2+2$ in your first step. – Sujaan Kunalan Apr 5 '14 at 17:26
• Oh woops thanks for the correction – TrueDefault Apr 5 '14 at 17:27
• No problem. Looks good now. (+1) – Sujaan Kunalan Apr 5 '14 at 17:36

HINT:

Just expand in Binomial Expansion to find a Quadratic Equation in $x^2$

Using the expansion $$\left(a+b\right)^n = \sum_{r=0}^n\binom{n}{r}a^rb^{n-r}$$

$$\require{cancel}{x^4+\cancel{4x^3}+6x^2+\cancel{4x}+1+x^4+-\cancel{4x^3}+6x^2-\cancel{4x}+1 = 16}$$

$$x^4+6x^2+1=8$$

$$t^2+6t-7=0\quad\text{with}\quad x^2=t$$

$$t=1,-7$$

Since $x^2=-7$ doesn't have solutions in reals, $x^2=1\implies x=\pm1$

Let denote by $P(X)$ the following polynomial:

$$P(X)=(X+1)^4+(X-1)^4-16.\quad\quad(1)$$

We immediately see that $X=1$ and $X=-1$ are roots of $P(X)$. Therefore, $P(X)$ can be written as:

$$P(X)=(X-1)(X+1)Q(X), \quad\quad\;\;\;\;\;\;(2)$$ where $Q(X)$ is a polynomial with degree $2$. Hence, $Q(X)=aX^2+bX+c$ for some $a$, $b$, and $c$.

Now, if we develop $P(X)$ for both $(1)$ and $(2)$ we get:

First, with $(1)$: $$P(X)=2X^4+12X^2-14.\quad\quad(3)$$ Second, with $(2)$: $$P(X)=(X-1)(X+1)(aX^2+bX+c)=aX^4+bX^3+(c-a)X^2-bX-c.\quad(4)$$ Hence, $(3)=(4)$

$$2X^4+12X^2-14=aX^4+bX^3+(c-a)X^2-bX-c.$$

Then, $$\begin{array}{lr} a=2\\ b=0\\ c=14 \end{array}$$

And finally,

$$P(X)=(X+1)(X-1)(2X^2+14)=2(X+1)(X-1)(X^2+7).$$

Now it is clear how to solve $P(X)=0$ (how to find the two other roots):

$$S_{\mathbb{R}}=\{-1, +1\}.$$ and $$S_{\mathbb{C}}=\{-1, +1, +i\sqrt{7}, -i\sqrt{7}\}.$$

Looking at $(x+1)^4 + (x-1)^4=16$, I recall that $16=2^4$. So I look at the equation trying to see if I could get a $2^4$ in there. This sugguests that we guess $x=1$ to make the left term a power of $2$.

Plugging in $x=1$,

$$(1+1)^4 + (1-1)^4 = 2^4 + 0^4 = 16$$

So $x=1$ is a solution.

This problem has a symmetry property. Notice that sending $x\rightarrow -x$ doesn't change the expression on the left.

$$(x+1)^4 + (x-1)^4 \rightarrow (-x+1)^4 + (-x-1)^4 = (x-1)^4+(x+1)^4$$

So we know that $x=-1$ will also be a solution.

Some thought should convince you the only possible solutions are between $-1$ and $1$ since any large values of $x$ will immediately make the expression on the left bigger than $16$.

Looking at the derivative of the expression on the left,

$$y' = 4(x+1)^3 + 4(x-1)^3,$$

we see that the derivative is $0$ at $x=0$ and monotonically increasing for $x > 0$ which means that there are no other solutions since our graph can only intersect the horizontal line $y=16$ once.

$(x+1)^4+(x-1)^4=16\\ \implies (x^4+4x^3+6x^2+4x+1)+(x^4-4x^3+6x^2-4x+1)-16=0\\ \implies (2x^4+12x^2+2)-16\\ \implies 2x^4+12x^2-14=0\\ \implies 2(x^4+6x^2-7)=0\\ \implies 2((x^2)^2+6(x^2)^1-7(x^2)^0)=0$

Now let $x^2=a$:

$2((a)^2+6(a)^1-7(a)^0)=0\\ \implies 2(a^2+6a-7)=0\\ \implies a^2+6a-7=0\\ \implies (a+7)(a-1)=0\\ \implies a=-7 \text{ or } a=1$.

Letting $a=x^2$ again, we get

$x^2=-7\text{ or } x^2=1$.

Hence the real solutions of $x$ are $\pm1$.